Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 13 - Exercises and Problems - Page 240: 73


27 degrees

Work Step by Step

We first find the maximum allowable velocity: $6 = \frac{mv^2}{r} + mg$ $6 = \frac{(.5)v^2}{.45} + (.5)(9.81)$ $v=.9927 \ m/s$ Thus, we find the height that it can go: $mgh = \frac{1}{2}mv^2 \\ h=\frac{v^2}{2g}=\frac{.9927^2}{2\times9.81}=.05$ Thus, we use trigonometry to find: $cos\theta = \frac{.4}{.45} \\ \theta = 27.26^{\circ}$
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